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Oduvanchick [21]
4 years ago
12

A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average

acceleration? 5 m/s2 10 m/s 10 m/s2 15 m/s
Physics
1 answer:
Elden [556K]4 years ago
4 0

The answer would be 10ms 2

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Astronomers know that the distance between the Earth and the Sun averages 1.50 x108 km. How can astronomers use the observed ste
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The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).

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Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.

Explanation:

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D = 1/p

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3 0
3 years ago
Which of the following statement is not true about machines?
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The answer to this question is D or the last one 

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4 years ago
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How would you describe brass since it ia used in weapons, pipes, intruments and ect? A compound, Alloy, Element l, Or molecule?
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5 0
3 years ago
A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch.A 8.2-V battery is connected
tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance  in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})

Put the value into the formula

I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})

I(t)=0.01925\ A

I(t) = 19.25\ mA

(B). We need to calculate the store energy in the inductor

Using formula of energy

E=\dfrac{1}{2}LI^2

Put the value into the formula

E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2

E=7.04\times10^{-6}\ J

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

8 0
3 years ago
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